Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int sum_row[105], sum_col[105], totalSum = 0;
int nr_row[105], nr_col[105];
int g[105][105];
pair<int, int> look_rows(int n, int m) {
  int max_sum = 0, min_nr = 0, ind_max = 0;
  for (int i = 1; i <= n; ++i) {
    if (nr_row[i] == m && sum_row[i] > max_sum) {
      max_sum = sum_row[i];
      ind_max = i;
      min_nr = g[i][1];
      for (int j = 2; j <= m; ++j) {
        min_nr = min(min_nr, g[i][j]);
      }
    }
  }
  return make_pair(ind_max, min_nr);
}
pair<int, int> look_cols(int n, int m) {
  int max_sum = 0, min_nr = 0, ind_max = 0;
  for (int i = 1; i <= m; ++i) {
    if (nr_col[i] == n && sum_col[i] > max_sum) {
      max_sum = sum_col[i];
      ind_max = i;
      min_nr = g[1][i];
      for (int j = 2; j <= n; ++j) {
        min_nr = min(min_nr, g[j][i]);
      }
    }
  }
  return make_pair(ind_max, min_nr);
}
string make_string(int nr) {
  string temp = "";
  while (nr > 0) {
    temp += (nr % 10) + '0';
    nr /= 10;
  }
  reverse(temp.begin(), temp.end());
  return temp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      cin >> g[i][j];
      sum_row[i] += g[i][j];
      sum_col[j] += g[i][j];
      totalSum += g[i][j];
      if (g[i][j] != 0) {
        ++nr_row[i];
        ++nr_col[j];
      }
    }
  }
  vector<string> sol;
  while (totalSum > 0) {
    int currentSum = totalSum;
    pair<int, int> max_sum_row_pos = look_rows(n, m);
    pair<int, int> max_sum_col_pos = look_cols(n, m);
    int pos_row = max_sum_row_pos.first;
    int pos_col = max_sum_col_pos.first;
    if (sum_row[pos_row] > sum_col[pos_col] && pos_row != 0) {
      int nr = max_sum_row_pos.second;
      nr = 1;
      for (int i = 1; i <= nr; ++i) {
        sol.push_back("row " + make_string(pos_row));
      }
      currentSum -= nr * m;
      sum_row[pos_row] -= nr * m;
      for (int i = 1; i <= m; ++i) {
        sum_col[i] -= nr;
        g[pos_row][i] -= nr;
        if (g[pos_row][i] == 0) {
          --nr_col[i];
          --nr_row[pos_row];
        }
      }
    } else if (pos_col != 0) {
      int nr = max_sum_col_pos.second;
      nr = 1;
      for (int i = 1; i <= nr; ++i) {
        sol.push_back("col " + make_string(pos_col));
      }
      currentSum -= nr * n;
      sum_col[pos_col] -= nr * n;
      for (int i = 1; i <= n; ++i) {
        sum_row[i] -= nr;
        g[i][pos_col] -= nr;
        if (g[i][pos_col] == 0) {
          --nr_row[i];
          --nr_col[pos_col];
        }
      }
    }
    if (currentSum == totalSum) {
      cout << -1;
      return 0;
    }
    totalSum = currentSum;
  }
  cout << sol.size() << "\n";
  for (string s : sol) {
    cout << s << "\n";
  }
  return 0;
}